Mathematica
Balkanica
––––––––––
New Series Vol. 16, 2002, Fasc. 1-4
Generalized Factorial Functions,
Numbers and Polynomials 1
Gradimir V. Milovanović ∗ , Aleksandar Petojević
∗∗
This paper is dedicated to Prof. Bl. Sendov on the occasion of his 70th anniversary
The generalized factorial functions and numbers and some classes of polynomials associated with them are considered. The recurrence relations, several representations, asymptotic
and other properties of such numbers and polynomials, as well as the corresponding generating
functions are investigated.
AMS Subj. Classification: Primary 11B83, 33E20; Secondary 33B15, 33B20
Key Words: generalized factorial functions, factorial numbers, factorial polynomials
1. Introduction and preliminaries
In 1971 Kurepa (see [10, 11]) defined so-called the left factorial !n by:
!0 = 0,
!n =
n−1
X
k=0
k! (n ∈ N)
and extended it to the complex half-plane Re z > 0 as
Z +∞ z
t − 1 −t
e dt.
K(z) =!z =
t−1
0
Such function can be also extended analytically to the whole complex plane by
K(z) = K(z + 1) − Γ(z + 1), where Γ(z) is the gamma function defined by
Z +∞
tz−1 e−t dt (Re z > 0)
and
zΓ(z) = Γ(z + 1).
Γ(z) =
0
1
This work was supported in part by the Serbian Ministry of Science, Technology and
Development under Grant # 2002: Applied Orthogonal Systems, Constructive Approximation
and Numerical Methods
114
G.V. Milovanović, A. Petojević
Kurepa [11] proved that K(z) is a meromorphic function with simple
poles at the points zk = −k (k ∈ N \ {2}). Slavić [23] found the representation
µ +∞
¶ X
+∞
1 X 1
π
+γ +
K(z) = − cot πz +
Γ(z − n),
e
e n=1 n!n
n=0
where γ is Euler’s constant. These formulas were mentioned also in the book
[14].
A number of problems and hypotheses, especially in number theory, were
posed by Kurepa and then considered by several mathematicians. For example,
Kurepa [10] asked if
gcd(!n, n!) = 2 (n = 2, 3, . . .),
where gcd(a, b) denotes the greatest common divisor of integers a and b. This
conjecture, known as the left factorial hypothesis (KH), is still an open problem
in number theory. There are several statements equivalent to KH. An equivalent
formulation of KH appears in the book [8, Problem B44],
!n 6≡ 0 (mod n)
for all n > 2.
Kurepa [12] also showed that KH can be reduced to primes so that KH is
equivalent to the following assertion
!p 6≡ 0 (mod p), for all primes p > 2.
For details see [18, 19, 20], as well as a recent survey written by Ivić and Mijajlović, [9].
Recently, Milovanović [16] defined and studied a sequence of the factorial
functions {Mm (z)}+∞
m=−1 , where M−1 (z) = Γ(z) and M0 (z) = K(z). Namely,
(1.1)
Mm (z) =
Z
0
+∞ z+m
t
− Qm (t, z) −t
e dt (Re z > −(m + 1)),
(t − 1)m+1
where the polynomials Qm (t; z), m = −1, 0, 1, 2, . . ., are given by
(1.2)
Since
(1.3)
Q−1 (t, z) = 0,
Qm (t, z) =
¶
m µ
X
m+z
(t − 1)k .
k
k=0
Mm (z) = Mm (z + 1) − Mm−1 (z + 1) (m ∈ N0 ),
Generalized Factorial Functions, . . .
115
similar to the gamma function, the functions z 7→ Mm (z), for each m ∈ N0 ,
can be extended analytically to the hole complex plane, starting from the corresponding analytic extension of the gamma function.
Suppose that we have analytic extensions for all functions z 7→ Mν (z),
ν < m. Let the function z 7→ Mm (z) be defined by (1.1) for z in the half-plane
Re z > −(m + 1). Using successively (1.3), we define at first Mm (z) for z in the
strip −(m + 2) < Re z < −(m + 1), then for z such that −(m + 3) < Re z <
−(m + 2), etc. In this way, we obtain the function Mm (z) in the hole complex
plane.
In the same paper [16] the numbers Mm (n) were introduced. For nonnegative integers n, m ∈ N0 we have
(1.4)
Mm (0) = 0,
Mm (n) =
n−1
X
i=0
¶
n−1 µ
(−1)i X
m+n
k!
.
i!
k+m+1
k=i
The numbers Mm (n) can be expressed in terms of the derangement numbers (cf. [22, p. 65], [5, p. 182], [16])
(1.5)
Sk = k!
k
X
(−1)ν
ν=0
in the form
(1.6)
Mm (n) =
ν!
(k ≥ 0)
n−1
Xµ
k=0
¶
m+n
Sk .
k+m+1
The numbers (1.5) satisfy the recurrence relation Sk = kSk−1 + (−1)k with
S0 = 1. Also, it is easy to prove that
Sk+2 = (k + 1)(Sk+1 + Sk )
(k ≥ 0).
Notice that S0 = 1, S1 = 0, S2 = 1, S3 = 2, S4 = 9, S5 = 44, . . . and
0 ≤ Sk < Sk+1 for k ∈ N. Their generating function is given by (see [24, p. 147]
and [17, Example 3])
+∞
X
xk
e−x
=
.
Sk
k!
1−x
k=0
In this paper we consider the factorial functions and numbers Mm (n),
some classes of polynomials associated with them, as well as some other related
problems.
The paper is organized as follows. In Section 2 we investigate the numbers
Mm (n). Generating functions for such numbers are given in Section 3. Factorial
116
G.V. Milovanović, A. Petojević
polynomials are defined and investigated in Section 4. Finally, some integral
representations of factorial functions Mm (z) are derived in Section 5.
2. The numbers Mm (n)
For a fixed m ∈ N, using (1.6) we obtain ([16])
Mm (1) = 1,
Mm (4) =
Mm (2) = m + 2,
1
Mm (3) = (m2 + 5m + 8),
2
1 3
(m + 9m2 + 32m + 60),
6
etc.
In general,
n!Mm (n + 1) =
n
X
A(m,n)
mν
ν
(A(m,n)
= 1).
n
ν=0
Thus, for a fixed n, we have Mm (n + 1) ∼ mn /n! as m → +∞.
Some values of the numbers Mm (n) are given in Table 1. The first row
(m = −1) represents factorials M−1 (n) = Γ(n) = (n − 1)!, and the second one
(m = 0) gives the Kurepa numbers (left facorials) M0 (n) = K(n) =!n.
Taking (1.3), i.e.,
(2.1)
Mν (n + 1) − Mν−1 (n + 1) = Mν (n)
for ν = 0, 1, . . . , m, we obtain
Mm (n + 1) − M−1 (n + 1) =
m
X
Mν (n).
ν=0
So, we get the following representation:
Lemma 2.1. For each n ∈ N,
Mm (n + 1) = n! +
m
X
Mν (n).
ν=0
Lemma 2.2. For each fixed ν ∈ N0 we have
(2.2)
Mν (n)
= 1.
n→+∞ Mν−1 (n)
lim
Generalized Factorial Functions, . . .
m\n
−1
0
1
2
3
4
5
1
1
1
1
1
1
1
1
2
1
2
3
4
5
6
7
3
2
4
7
11
16
22
29
117
4
6
10
17
28
44
66
95
5
24
34
51
79
123
189
284
6
120
154
205
284
407
596
880
7
720
874
1079
1363
1770
2366
3246
8
5040
5914
6993
8356
10126
12492
15738
Table 1: The numbers Mm (n) for m = −1, 0, 1, . . . , 5 and n = 1, 2, . . . , 8
P r o o f. First, we note that all sequences {Mm (n)}+∞
n=1 (m ≥ −1) are
increasing, as well as
M0 (n)
n→+∞ M−1 (n)
lim
=
=
K(n)
K(n) − K(n − 1)
= lim
n→+∞ Γ(n)
n→+∞ Γ(n) − Γ(n − 1)
lim
lim
n→+∞
Γ(n)
= 1.
Γ(n) − Γ(n − 1)
Using Stolz’ theorem and relation (2.1) we get
Mν (n)
n→+∞ Mν−1 (n)
lim
=
=
Since
Mν (n) − Mν (n − 1)
n→+∞ Mν−1 (n) − Mν−1 (n − 1)
lim
lim
n→+∞
Mν−1 (n)
M0 (n)
= · · · = lim
= 1.
n→+∞ M−1 (n)
Mν−2 (n)
M0 (n) M1 (n)
Mm (n)
Mm (n)
=
·
···
,
M−1 (n)
M−1 (n) M0 (n)
Mm−1 (n)
using (2.2) we get the following result:
Theorem 2.3. For each m ∈ N0 we have
(2.3)
lim
n→+∞
Mm (n)
=1
(n − 1)!
and
lim
n→+∞
Mm (n)
= 0.
n!
As we can see the asymptotic relation Mm (n + 1) ∼ nMm (n) (n → +∞)
holds. Furthermore, for a sufficiently large n, we can prove an inequality:
118
G.V. Milovanović, A. Petojević
Theorem 2.4. For each m ∈ N0 there exists an integer nm ∈ N such
that for every n ≥ nm the following inequality
Mm (n + 1) ≤ nMm (n)
(2.4)
holds.
P r o o f. First, we note that for m = −1, the inequality (2.4) reduces to
the well-known functional equation for the gamma function, Γ(n + 1) = nΓ(n),
n ≥ 1.
For m = 0, inequality (2.4) reduces to !(n + 1) ≤ n(!n), which is not true
for n = 1. But, it can be proved for each n ≥ n0 = 2. Indeed, for n = 2, (2.4)
becomes an equality !3 = 2(!2), i.e., 0! + 1! + 2! = 2(0! + 1!) = 4. Suppose now
that (2.4) holds for some n = k ≥ 2, i.e.,
M0 (k + 1) ≤ kM0 (k).
Adding M−1 (k + 2) = Γ(k + 2) = (k + 1)Γ(k + 1) to both sides in the previous
inequality, we get
M0 (k + 1) + M−1 (k + 2) ≤ kM0 (k) + (k + 1)M−1 (k + 1)
= (k + 1)(M0 (k) + M−1 (k + 1)) − M0 (k).
Using recurrence relation (2.1) for ν = 0 and a fact that M0 (k) > 0, we
obtain
M0 (k + 2) ≤ (k + 1)M0 (k + 1).
Notice that Mm (4) ≤ 3Mm (3) for m = 1, 2, 3, 4 (see Table 1), so that
nm = 3 for such values of m.
For m ≥ 5 we can prove inequality (2.4) for n = m, i.e., Mm (m + 1) ≤
mMm (m). This means that we can take nm = m for m ≥ 5. According to (1.3)
and (1.6) this inequality can be represented in the following equivalent forms:
Mm−1 (m + 1) ≤ (m − 1)Mm (m)
and
¶
m µ
m−1
X
X µ 2m ¶
2m
Sk ≤ (m − 1)
Sk .
k+m
k+m+1
k=0
k=0
+ (−1)k ,
Because of Sk = kSk−1
the last inequality reduces to
¶
¶
m µ
m µ
X
X
2m
2m
k
(−1) ≤
(m − 1 − k)Sk−1 .
k+m
k+m
k=0
k=1
Generalized Factorial Functions, . . .
119
Since the sum on the left-hand side of the previous inequality is equal to
(cf. [21, p. 607]), we get
¶
¶ X
µ
m µ
2m
2m − 1
(m − 1 − k)Sk−1 ,
≤
k+m
m
¡2m−1¢
m−1
k=1
i.e.,
m−3
Xµ
(2.5)
k=2
¶
2m
(m − 1 − k)Sk−1 + Am + Bm ≥ 0,
k+m
where
µ
¶
µ
¶ µ
¶
2m
2m − 1
2m2 − 5m − 1 2m − 1
−
=
Am = (m − 2)
m+1
m
m+1
m
and
Bm
¶
µ ¶
∙
¸
2m
2m
Sm−1
Sm−3 −
Sm−1 = Sm−3 m(2m − 1) −
=
.
2m − 2
2m
Sm−3
µ
Since Sm−1 /Sm−3 = (m − 2)(m − 1 + (−1)m /Sm−3 ) we have that
∙
¸
(−1)m
6
−
>0
Bm = (m − 2)Sm−3 m + 4 +
m−2
Sm−3
for m ≥ 5. Notice also that Am > 0 for m ≥ 3.
Since the first term in (2.5) is equal to zero and others are positive, we
conclude that the inequality (2.5) is true for each m ≥ 5. Thus, this proves the
existence of the numbers nm for each m.
The proof of (2.4) for m ∈ N and n ≥ nm can be given by induction in
m. Namely, supposing that (2.4) holds for each ν < m ∈ N and n ≥ nν (n1 =
· · · = n4 = 3, nν = ν for ν > 4) we prove the inequality Mm (n + 1) ≤ nMm (n)
for n ≥ nm = m. In order to do this we apply induction in n, in the same way
as for m = 0.
Since (2.4) holds for m = 0 and n ≥ n0 = 2, it means that (2.4) holds
for each m ∈ N0 and n ≥ nm .
R e m a r k 2.5. Theorem 2.4 establishes only the existence of the numbers nm . The minimal values of mn can be expressed in the following way, if we
define
aν = ν 2 + ν − 1,
bν = ν 2 + 3ν,
Iν = {m ∈ Z : m ∈ [aν , bν ]},
120
G.V. Milovanović, A. Petojević
for each ν ∈ N0 . Notice that aν+1 = bν +1, I0 = {−1, 0}, as well as I1 ∪I2 ∪ · · · =
N. Then for each ν ∈ N and m ∈ Iν we have nm = ν + 2.
For example, for ν = 1, i.e., m ∈ {1, 2, 3, 4}, we have nm = 3. It was
noted in the proof of the previous theorem. For m ∈ {5, 6, 7, 8, 9, 10} (ν = 2)
we have nm = 4, etc. Notice that nm < m.
3. Generating functions for the numbers Mm (n)
Definition 3.1. Let m ∈ N0 . The exponential generating function of
the sequence {Mm (n)}+∞
n=0 is given by
(3.1)
gm (x) =
+∞
X
Mm (n)
n=0
xn
.
n!
According to Theorem 2.3, the expansion (3.1) converges in the unit
circle |x| < 1. Notice also that gm (0) = 0.
R e m a r k 3.2. Because of Mm (0) = 0 we have
(3.2)
gm (x) =
+∞
X
Mm (n)
n=1
xn
.
n!
This modification in previous definition enables us to define the exponential
+∞
generating function of the sequence {Γ(n)}+∞
n=1 , i.e., {(n − 1)!}n=1 . Thus, for
m = −1, (3.2) reduces to
(3.3)
g−1 (x) =
+∞
X
n=1
Theorem 3.3.
following relation
(3.4)
+∞
Γ(n)
1
xn X xn
=
= log
n!
n
1−x
n=1
The generating functions gm (m = 0, 1, . . .) satisfy the
x
gm+1 (x) = gm (x) + e
Z
x
e−t gm (t) dt,
0
where
(3.5)
(|x| < 1).
g0 (x) = ex−1 (Ei (1) − Ei (1 − x)),
and Ei (x) is the exponential integral defined by
Z x t
e
Ei (x) = v.p.
(3.6)
dt
(x > 0).
−∞ t
m ≥ 0,
Generalized Factorial Functions, . . .
121
P r o o f. Let |x| < 1 and gm be defined by (3.1). Then
0
(x) =
gm
+∞
X
+∞
Mm (n)
n=1
and
0
0
(x) − gm
(x) =
gm+1
i.e.,
+∞
X
n=0
X
xn−1
xn
=
Mm (n + 1)
(n − 1)! n=0
n!
(Mm+1 (n + 1) − Mm (n + 1))
0
0
gm+1
(x) − gm
(x) =
+∞
X
Mm+1 (n)
n=0
xn
,
n!
xn
= gm+1 (x).
n!
Integrating this differential equation, we obtain
Z x
x
0
gm+1 (x) = e
(3.7)
e−t gm
(t) dt.
0
Finally, an integration by parts gives
x
gm+1 (x) = gm (x) + e
Z
x
e−t gm (t) dt.
0
According to (3.7) and (3.3), for m = 0 we have
Z x −t
Z 1 u−1
e
e
dt = ex
du.
g0 (x) = ex
1
−
t
0
1−x u
Using the exponential integral Ei (x) defined by (3.6) (see [1, p. 228]), the previous formula becomes
g0 (x) = ex−1 (Ei (1) − Ei (1 − x))
(|x| < 1),
i.e., (3.5).
R e m a r k 3.4. The generating function for left factorial in the form
- . Cvijović.
(3.5) was obtained by D
In order to find an explicit expression for gm (x) we denote its Laplace
transform by Gm (s). Then, from (3.4) it follows
¶
µ
1
1
Gm (s) = 1 +
Gm (s),
Gm+1 (s) = Gm (s) +
s−1
s−1
122
G.V. Milovanović, A. Petojević
i.e.,
µ
Gm (s) = 1 +
1
s−1
¶m
G0 (s) =
m µ ¶
X
m
k=0
k
1
· G0 (s).
(s − 1)k
The inverse transform gives
(3.8)
gm (x) = g0 (x) +
m µ ¶
X
m
k
k=1
1
(k − 1)!
Z
0
x
ex−t (x − t)k−1 g0 (t) dt.
Using integration by parts in the integral convolutions on the right hand side in
(3.8) yields
Z
x
x−t
e
0
k−1
(x − t)
g0 (t) dt =
=
Since
Z
0
x
−t
ν−1
e (1 − t)
dt =
Z
x
dt
1
−t
0
¶
µ
Z x
k
ex X k
k−ν
(x − 1)
e−t (1 − t)ν−1 dt.
k ν=0 ν
0
1
k
(
ex−t (x − t)k
e−x g0 (x),
Pν−1 (0) − e−x Pν−1 (x), if ν ≥ 1,
where
ν
Pν (x) = (−1) ν!
(3.9)
if ν = 0,
ν
X
(x − 1)j
j=0
j!
,
we get
Z
0
x
x−t
e
k−1
(x − t)
g0 (t) dt =
k µ ¶
(x − 1)k
ex X k
(x − 1)k−ν Pν−1 (0)
g0 (x) +
ν
k
k
ν=1
−
k µ ¶
1X k
(x − 1)k−ν Pν−1 (x).
k ν=1 ν
According to this equality, (3.8) and (3.9) we have the following result:
Theorem 3.5. For each m ∈ N0 the generating function x 7→ gm (x) is
given by
1
(Am (x)g0 (x) + Bm (x)ex − Cm (x)),
(3.10)
gm (x) =
m!
Generalized Factorial Functions, . . .
123
where Am (x), Bm (x), and Cm (x) are polynomials determined by
m µ ¶
X
Am (x)
m (x − 1)k
=
,
m!
k!
k
k=0
Bm (x)
m!
Cm (x)
m!
respectively.
=
m−1
X
ν=0
=
m−1
X
j=0
⎛
m−ν
Xµ
⎝
k=1
m
k+ν
¶
k−1
(−1)k−1 X
k
j=0
(−1)j
⎞
ν
⎠ (x − 1) ,
j!
ν!
⎞
⎛
µ ¶
µ ¶ X
j
m
k−1 m
j
X
(−1)
j
⎠ (x − 1) ,
⎝ (−1)ν
k−ν
k
j!
ν
ν=0
k=j+1
The polynomials Am (x), Bm (x), and Cm (x) for 1 ≤ m ≤ 6 are presented
in Table 2.
m
1
2
3
4
5
6
Am (x)
x
x2 + 2x − 1
x3 + 6x2 + 3x − 4
x4 + 12x3 + 30x2 − 4x − 15
x5 + 20x4 + 110x3 + 140x2
− 95x − 56
6
x + 30x5 + 285x4 + 940x3
+555x2 − 906x − 185
Bm (x)
1
2x + 2
3x2 + 12x + 4
4x3 + 36x2 + 64x + 6
5x4 + 80x3 + 340x2
+ 350x − 16
5
6x + 150x4 + 1160x3
+3090x2 + 2004x− 310
Cm (x)
1
x+2
x2 + 6x + 4
x3 + 12x2 + 31x + 6
x4 + 20x3 + 111x2
+158x − 16
5
x + 30x4 + 286x3
+968x2 + 789x − 310
Table 2: The polynomials Am (x), Bm (x), and Cm (x) in (3.10) for m =
1, 2, 3, 4, 5, 6
R e m a r k 3.6.
Starting from A0 (x) = 1, B0 (x) = C0 (x) = 0, the
polynomials Am (x), Bm (x), and Cm (x) can be calculated recursively by
µ
¶
Z x
Am+1 (x) = (m + 1) Am (x) +
Am (t) dt ,
1
µ
¶
Z x
Bm+1 (x) = (m + 1) Bm (x) +
Bm (t) dt + βm+1 ,
0
Z
Cm+1 (x) = ex e−x ϕm (x) dx,
124
G.V. Milovanović, A. Petojević
where ϕm (x) is a polynomial of degree m defined by
µ
¶
Z x
1
0
ϕm (x) = (m + 1) Cm
(x) −
Am (t) dt ,
x−1 1
and βm+1 = Cm+1 (0) − (m + 1)Cm (0).
R e m a r k 3.7. It is clear that Am (x) > 0 for x ≥ 1 and Am (1) = m!.
Also, Bm (0) = Cm (0). It can be proved that polynomials Am (x) have only real
zeros distributed in (−∞, 1). Furthermore, the zeros of Am (x) and Am+1 (x)
mutually separate each other.
4. The factorial polynomials
Definition 4.1.
are defined by
(m)
Let m ∈ N0 . The factorial polynomials {Kn (t)}n∈N
Gm (t, x) = ext gm (x) =
+∞
X
Kn(m) (t)
n=1
xn
,
n!
where gm is defined by (3.1) and given by (3.10).
Using (3.1) and the numbers Mm (k) it is easy to prove the following
explicit representation of the factorial polynomials:
Theorem 4.2. For each m ∈ N0 and n ∈ N we have
Kn(m) (t)
n µ ¶
X
n
=
Mm (k)tn−k .
k
k=1
For example, for m = 0 and n ≤ 7 we have
(0)
K1 (t) = 1,
(0)
K2 (t) = 2t + 2,
(0)
K3 (t) = 3t2 + 6t + 4,
(0)
K4 (t) = 4t3 + 12t2 + 16t + 10,
(0)
K5 (t) = 5t4 + 20t3 + 40t2 + 50t + 34,
(0)
K6 (t) = 6t5 + 30t4 + 80t3 + 150t2 + 204t + 154,
(0)
K7 (t) = 7t6 + 42t5 + 140t4 + 350t3 + 714t2 + 1078t + 874.
Generalized Factorial Functions, . . .
125
Since
n µ ¶
X
d (m)
n
Kn (t) =
(n − k)Mm (k)tn−1−k
dt
k
k=1
n−1
Xµ
=
k=1
¶
n−1
Mm (k)tn−1−k ,
k
we have the following differentiation formula
d (m)
(m)
K (t) = nKn−1 (t).
dt n
Also, we have
dν (m)
(m)
K (t) = n(n − 1) · · · (n − ν + 1)Kn−ν (t)
dtν n
(0 < ν < n).
(m)
Expanding Kn (t + s) in Taylor series and using the previous formula
we obtain
+∞
n−1
X
X µn¶ (m)
1 dν (m)
ν
Kn(m) (t + s) =
Kn−ν (t)sn ,
K
(t)s
=
n
ν
ν!
dt
ν
ν=0
ν=0
i.e.,
Kn(m) (t + s)
It is clear that
n µ ¶
X
n
Kν(m) (t)sn−ν .
=
ν
ν=1
dν (m)
K (t) > 0
dtν n
(0 ≤ ν < n)
(m)
for each t ≥ 0. Therefore, the polynomials Kn (t) have no positive real zeros.
A simple representation of these polynomials in terms of zeros can be
done:
(m)
(m)
Theorem
4.3. Let τν (ν = 1, . . . , n) be zeros of Kn+1 (t), i.e., Kn+1 (t) =
Qn
(n + 1) ν=1 (t − τν ). Then
(m)
Kn(m) (t) =
n
n
n
1 X Kn+1 (t) X Y
=
(t − τν ).
n+1
t − τk
ν=1
k=1
k=1
ν6=k
126
G.V. Milovanović, A. Petojević
5. The factorial functions Mm (z)
In this section we consider again the factorial functions Mm (z) defined
by (1.1) and (1.2). First, we put
Ym (t, z) =
tz+m − Qm (t, z)
(t − 1)m+1
(Re z > −(m + 1)).
Using the binomial series
t
z+m
m+z
= (1 + t − 1)
¶
+∞ µ
X
m+z
(t − 1)k
=
k
k=0
(|t − 1| < 1),
we see that, for 0 < t < 2,
¶
+∞ µ
X
m+z
Ym (t, z) =
(t − 1)k .
k+m+1
k=0
The function Ym (t, z) can be expressed in terms of the hypergeometric function
2 F1 , defined by
+∞
X
(a)k (b)k xk
F
(a,
b,
c;
x)
=
·
2 1
(c)k
k!
k=0
for |x| < 1, and by continuation elsewhere. It is well-known that
Γ(c)
2 F1 (a, b, c; x) =
Γ(b)Γ(c − b)
Z
1
0
ξ b−1 (1 − ξ)c−b−1 (1 − xξ)−a dξ
in the x plane cut along the real axis from 1 to ∞, if Re c > Re b > 0 (cf. [2,
p. 65]). Here it is understood that arg ξ = arg(1 − ξ) = 0 and (1 − xξ)−a has its
principal value.
According to (1.1) we note that Mm (z) can be interpreted as the Laplace
transform of the function t 7→ Ym (t, z) at the point s = 1. Therefore, we put
(5.1)
Fm (s, z) = L[Ym (t, z)] =
Z
+∞
Gm (t, z)e−st dt,
0
where z is a complex parameter such that Re z > −(m + 1), and Mm (z) =
Fm (1, z).
Generalized Factorial Functions, . . .
127
Theorem 5.1. For Re z > −(m+1), the factorial functions z 7→ Mm (z)
have the integral representation
Z
³ 1 − ξ´
z(z + 1) · · · (z + m) 1 z−1
dξ,
ξ (1 − ξ)m e(1−ξ)/ξ Γ z,
Mm (z) =
m!
ξ
0
where Γ(z, x) is the incomplete gamma function defined by
Z +∞
Γ(z, x) =
(5.2)
tz−1 e−t dt.
x
P r o o f. Since
and
(k + m + 1)! = (m + 1)!(m + 2)k
(1 − z)k =
(−1)k Γ(z)
,
Γ(z − k)
we have
µ
¶
Γ(m + z + 1) (1 − z)k (−1)k
m+z
Γ(m + z + 1)
=
·
,
=
Γ(z − k)(k + m + 1)!
Γ(z)(m + 1)!
(m + 2)k
k+m+1
so that
+∞
Ym (t, z) =
Γ(m + z + 1) X (1 − z)k (1)k (1 − t)k
·
Γ(z)(m + 1)!
(m + 2)k
k!
k=0
=
Γ(m + z + 1)
2 F1 (1 − z, 1, m + 2; 1 − t),
Γ(z)(m + 1)!
or, by continuation,
Γ(m + z + 1)
Ym (t, z) =
Γ(z)m!
Z
1
0
(1 − ξ)m (1 − (1 − t)ξ)z−1 dξ.
According to (5.1) we have
Fm (s, z) =
=
z(z + 1) · · · (z + m)
m!
z(z + 1) · · · (z + m)
m!
where α = (1 − ξ)/ξ.
Z
+∞
e−st
0
Z
0
Z
0
1
ξ
z−1
1
(1 − ξ)m ξ z−1 (t + α)z−1 dξ dt
m
(1 − ξ)
Z
0
+∞
e−st (t + α)z−1 dt dξ
128
G.V. Milovanović, A. Petojević
Since
eαs Γ(z, αs)
(Re s > 0),
sz
where Γ(z, x) is the incomplete gamma function defined by (5.2), we get
Z
z(z + 1) · · · (z + m) 1 z−1
Fm (s, z) =
ξ (1 − ξ)m eαs Γ(z, αs) dξ.
m! sz
0
L[(t + α)z−1 ] =
Finally, for s = 1 we obtain the result of the theorem.
Changing variables (1 − ξ)/ξ = x we get an alternative form of the
previous theorem:
Corollary 5.2.
For Re z > −(m + 1), the factorial functions z 7→
Mm (z) have the integral representation
Z
z(z + 1) · · · (z + m) +∞ xm ex Γ(z, x)
Mm (z) =
dx.
m!
(x + 1)z+m+1
0
In a special case when m = 0, we get the integral representation of the
Kurepa’s function
Z 1
Z +∞ x
³ 1 − ξ´
e Γ(z, x)
ξ z−1 e(1−ξ)/ξ Γ z,
dx,
K(z) = z
dξ = z
ξ
(x + 1)z+1
0
0
which holds for Re z > −1. In 1995 one of us [15] derived the Chebyshev
expansion for K(1 + z) and 1/K(1 + z), as well as the power series expansion
of K(a + z), a ≥ 0, and determined numerical values of their coefficients bν (a)
for a = 0 and a = 1. Using an asymptotic behaviour of bν (a), when ν → ∞, a
transformation of series with much faster convergence was obtained. For similar
expansions of the gamma function see e.g. Davis [6], Luke [13], Fransén and
Wrigge [7], and Bohman and Fröberg [4].
R e m a r k 5.3.
The function x 7→ x−z ex Γ(z, x) can be expanded in
continuous fractions (cf. [3, Chapter 9])
1
x−z ex Γ(z, x) =
.
1−z
x+
1
1+
2−z
x+
1 + ···
R e m a r k 5.4.
0, 1, . . . , m.
The function z 7→ Mm (z) has zeros at z = −n, n =
Generalized Factorial Functions, . . .
129
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∗
University of Niš
Fac. of Electronic Engineering, Dept. of Mathematics
P.O. Box 73, 18000 Niš, YUGOSLAVIA
e-mail: [email protected]
∗∗
University of Novi Sad
Fac. of Civil Engineering, Dept. of Mathematics
Kozaracka 2a, 24000 Subotica, YUGOSLAVIA
e-mail: [email protected]
Received: 17.03.2002